Q.Show that: $(\cosh x + \sinh x)^n = \cosh(nx) + \sinh(nx)$, for any $n \in R$.

Factor as a difference of squares; the identities $\cosh^2 x-\sinh^2 x=1$ and $\cosh^2 x+\sinh^2 x=\cosh 2x$ finish it.

Write the left side as a difference of two squares:

$\cosh^4 x-\sinh^4 x=(\cosh^2 x-\sinh^2 x)(\cosh^2 x+\sinh^2 x).$

Use the fundamental identity $\cosh^2 x-\sinh^2 x=1$, so the first factor is $1$: …

Rewrite $\tanh^2\frac{x}{2}$ in terms of $\cosh x$, substitute $\cosh x=\sec\theta$, and simplify to the half-angle form of $\tan^2\frac{\theta}{2}$.

Given $\cosh x = \sec\theta$.

Step 1. Use the half-argument identity for hyperbolic tangent:

$\tanh^2\dfrac{x}{2} = \dfrac{\cosh x - 1}{\cosh x + 1}$

Step 2. Substitute $\cosh x=\sec\theta$:

$\tanh^2\dfrac{x}{2} = \dfrac{\sec\theta - 1}{\sec\theta + 1} = \dfrac{\frac{1}{\cos\theta}-1}{\frac{1}{\cos\theta}+1} = \dfrac{1-\cos\theta}{1+\cos\theta}$

Step 3. Recall the ordinary half-angle identity: …

Find $\cosh x$ from $\cosh^2x-\sinh^2x=1$, then use $\sinh 2x = 2\sinh x\cosh x$.

Given $\sinh x = \dfrac{3}{4}$.

Step 1. Use $\cosh^2 x - \sinh^2 x = 1$:

$\cosh^2 x = 1+\sinh^2 x = 1+\dfrac{9}{16} = \dfrac{25}{16} \Rightarrow \cosh x = \dfrac{5}{4}\ (\cosh x>0\text{ always})$

…

Factor the difference of squares, then use the two standard hyperbolic identities $\cosh^2x-\sinh^2x=1$ and $\cosh^2x+\sinh^2x=\cosh2x$.

Step 1. Factor as a difference of squares:

$\cosh^4x-\sinh^4x=(\cosh^2x-\sinh^2x)(\cosh^2x+\sinh^2x)$

Step 2. Use the fundamental hyperbolic identity $\cosh^2x-\sinh^2x=1$:

$\cosh^4x-\sinh^4x=(1)(\cosh^2x+\sinh^2x)=\cosh^2x+\sinh^2x$

…

cosh x + sinh x equals eˣ by definition, so raising to the power n directly gives eⁿˣ, which equals cosh(nx)+sinh(nx).

Recall the definitions: $\cosh x = \dfrac{e^x+e^{-x}}{2}$, $\sinh x=\dfrac{e^x-e^{-x}}{2}$

Adding: $\cosh x+\sinh x = \dfrac{e^x+e^{-x}}{2}+\dfrac{e^x-e^{-x}}{2} = \dfrac{2e^x}{2} = e^x$

Therefore:

$(\cosh x+\sinh x)^n = (e^x)^n = e^{nx}$

But by the same definitions applied to nx:

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